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Math Help - Linear Algebra: Eigenvalues question

  1. #1
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    Linear Algebra: Eigenvalues question

    Given a matrix A over field F of order 5x5, with tr(A)=0, and p(A)=1, find the eigenvalues of A.

    If the question would have been over C/R I would have found an answer, but this F field thing is too much for me...

    Thanks!
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  2. #2
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    Quote Originally Posted by openustudent View Post
    Given a matrix A over field F of order 5x5, with tr(A)=0, and p(A)=1, find the eigenvalues of A.

    If the question would have been over C/R I would have found an answer, but this F field thing is too much for me...

    Thanks!
    It will help if you tell us what your notation means. what does p(A) = 1 mean?
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  3. #3
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    p(A) is the rank of the matrix A (in our case - 1) and tr(A) is the trace of the matrix A.
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    Quote Originally Posted by openustudent View Post
    Given a matrix A over field F of order 5x5, with tr(A)=0, and p(A)=1, find the eigenvalues of A.

    If the question would have been over C/R I would have found an answer, but this F field thing is too much for me...

    Thanks!
    Since the rank is 1, there exists vectors x and y such that A = xy^T(Prove this!)

    This matrix has atmost one non zero eigenvalue. Clearly x is an eigen vector corresponding to the non-zero(if it is...) eigenvalue x^T y.

    But observe that \text{Tr} (xy^T) = x^Ty, thus all eigenvalues of A are 0.
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  5. #5
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    Thank you for your reply!
    There are two statements though that I did not understand:

    This matrix has atmost one non zero eigenvalue
    How come ?

    Clearly is an eigen vector corresponding to the non-zero(if it is...) eigenvalue .
    Why is this?
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  6. #6
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    Quote Originally Posted by openustudent View Post
    How come ?
    Why is this?
    Answer to second question:
    Observe that Ax = (xy^T)x = x(y^Tx) = (y^Tx)x. If you still cant see it, observe that y^Tx is a scalar.

    Answer to question 1:
    We will show that any eigenvector corresponding to a non-zero eigenvalue is a scalar multiple of x.

    Let us say it had an eigenvector z, corresponding to a eigen value \alpha ,then Az = \alpha z \implies (xy^T)z = \alpha z \implies (y^T z) x = \alpha z , Thus if \alpha \neq 0, z is in the span of x.
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  7. #7
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    Thanks!
    I got it...
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